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The Existence of a Non-Lebesgue Measurable Set

So far we have noted that many subsets of $\mathbb{R}$ are Lebesgue measurable, so the next question we may ask is if every subset of $\mathbb{R}$ is Lebesgue measurable. The answer is no. We will now demonstrate the existence of a non-Lebesgue measurable set.

Note that $\sim$ is an equivalence relation on $[0, 1]$. To see this, we note that for all $x \in [0, 1]$ we have that $x - x = 0 \in \mathbb{Q}$ so $x \sim x$. This shows that $\sim$ is reflexive. Now let $x, y \in [0, 1]$ and suppose that $x \sim y$. Then $(x - y) \in \mathbb{Q}$. So clearly $(y - x) \in \mathbb{Q}$, and $y \sim x$. This shows that $\sim$ is symmetric. Now let $x, y, z \in [0, 1]$ and suppose that $x \sim y$ and $y \sim z$ then $(x - y) \in \mathbb{Q}$ and $(y - z) \in \mathbb{Q}$. So their sum $(x - z) \in \mathbb{Q}$ and $x \sim z$. This show that $\sim$ is transitive.

Since $\sim$ is an equivalence relation on $[0, 1]$ we can partition $[0, 1]$ into equivalence classes of $\sim$ that are mutually disjoint. By the Axiom of Choice, we can choose a representative of each equivalence class. Let $N$ be a set of such representative from these equivalence classes.

Now let $\{ r_1, r_2, ... \}$ be an enumeration of $\mathbb{Q} \cap [-1, 1]$. Consider the collection of sets: